# labelparttiny1 of 1Exercise 1. Let A E Kn over K. The column space of A is the subspace ofKm spanned by the columns of A, in short Col(Ais the ith column of A, similarly for the rouw space Row(A). The null space of A isNul(A) E K" | Ar = 0}C K"For this problem, K = Z2. Let f : K5A) where ASpan(AK4 by f()Ar for E Kwhere111111A=10001EKXS1100 1101(c.1 Apply row operations to A to get a reduced row Echelon form A(c.2) Use Problem (b) in HW2 to claim that Row(A Row(A'), then get a basisfor the row space Row(A) HW2 b is provided at the end of the question(c.3) Find a basis for Nul(A) Nul (A') (you should know that the two spacesare the same)(c.4 Using the basis in (c.3) to express each column of A a combination of someindependent columns of A, hence get a basis for the column space (A)(c.5) Is finjective or onto? What's the dimension for the kernel space and imagespace of f?(c.6) (About dual space) Note that each vector vE K5 defines a linear functionalKs K by v*(r (v, a)- ΣΣ 0π, ε κ .(A) and define W= {v* € (K5)* : v* (u) = 0 for all u E W}. Get afor K Letbasis for W using (c.3)For your reference, this was HW2, Problem b Let E be any vector spaceover K and u;}ieI any family of vectors in E (the family could be infinite). Letus, be any two vectors in the family (st) and let a E K be arbitrary. Showthat Span(u, u}iel,) = Span(u, + aut, {u;}ie!,)where I\I{s}. This means that one can add any multiple of any vector to anyother vector in the family, and the spanning space does not change. Hence deducethat Span(w, fu}ieh) = Span(u, {u}ien)where w is a finite combination of the vectors in the family that contains u, witha nonzero coefficient. (Here one should make connection to row operations of amatrix and how the row space changes.)

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Please, I need a detailed and easy to understand solution to (c.4)

Step 1

Obtain the basis o...

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